In the Leetcode Unique Paths II problem solution You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m – 1][n – 1]). The robot can only move either down or right at any point in time.
An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.
Return the number of possible unique paths that the robot can take to reach the bottom-right corner.
The testcases are generated so that the answer will be less than or equal to 2 * 109.
Example 1:
Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2
Explanation: There is one obstacle in the middle of the 3×3 grid above.
There are two ways to reach the bottom-right corner:
- Right -> Right -> Down -> Down
- Down -> Down -> Right -> Right
Example 2:
Input: obstacleGrid = [[0,1],[0,0]]
Output: 1
Constraints:
- m == obstacleGrid.length
- n == obstacleGrid[i].length
- 1 <= m, n <= 100
- obstacleGrid[i][j] is 0 or 1.
Solution in C Programming
int uniquePathsWithObstacles(int** arr, int n, int* m)
{
for (int i = 0; i < n; i++)
{
for (int j = 0; j < *m; j++)
{
if (arr[i][j] == 1)
arr[i][j] = 0;
else if (j == 0 && i == 0)
arr[i][j] = !arr[i][j];
else if (i == 0)
arr[i][j] = arr[i][j - 1];
else if (j == 0)
arr[i][j] = arr[i - 1][j];
else
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
return (arr[n - 1][*m - 1]);
}Solution in C++ Programming
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
if(obstacleGrid[0][0] == 1)return 0;
int n = obstacleGrid.size();
int m = obstacleGrid[0].size();
vector<vector<int>>dp(n,(vector<int>(m,1)));
for(int i = 0;i < n;i++){
if(obstacleGrid[i][0] == 1){
while(i < n){
dp[i][0] = 0;
i++;
}
}
}
for(int j = 0;j < m;j++){
if(obstacleGrid[0][j] == 1){
while(j < m){
dp[0][j] = 0;
j++;
}
}
}
for(int i = 1;i < n;i++){
for(int j = 1;j < m;j++){
if(obstacleGrid[i][j] == 1){
dp[i][j] = 0;
}
else{
int up = 0,left = 0;
if(i > 0) up = dp[i-1][j];
if(j > 0) left = dp[i][j-1];
dp[i][j] = left + up;
}
}
}
for(int i = 0;i < n;i++){
for(int j = 0;j < m;j++){
cout << dp[i][j] << " ";
}
cout << endl;
}
return dp[n-1][m-1];
}
};Solution in Java Programming
class Solution {
// Using tabulation
public int uniquePathsWithObstacles(int[][] grid) {
int n = grid.length;
int m = grid[0].length;
int[][] dp = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) {
dp[i][j] = 0;
} else if (i == 0 && j == 0)
dp[i][j] = 1;
else {
int up = 0, left = 0;
if (i > 0) up = dp[i - 1][j];
if (j > 0) left = dp[i][j - 1];
dp[i][j] = up + left;
}
}
}
return dp[n - 1][m - 1];
}
}Solution in Python Programming
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
:type obstacleGrid: List[List[int]]
:rtype: int
"""
m = len(obstacleGrid)
n = len(obstacleGrid[0])
dp = [[0]*n for _ in range(m)]
dp[0][0] = 1 - obstacleGrid[0][0]
for j in range(1, n):
if obstacleGrid[0][j]!= 1:
dp[0][j] = dp[0][j-1]
else:
dp[0][j] = 0
for i in range(1, m):
if obstacleGrid[i][0] != 1:
dp[i][0] = dp[i-1][0]
else:
dp[i][0] = 0
for i in range(1, m):
for j in range(1, n):
if obstacleGrid[i][j]!= 1:
dp[i][j] = dp[i-1][j] + dp[i][j-1]
else:
dp[i][j] = 0
return dp[-1][-1]Solution in C# Programming
public class Solution {
public int UniquePathsWithObstacles(int[][] obstacleGrid) {
/*
At any cell in the grid, we can only get there by going down one or to the right one.
<=> grid[i][j] = grid[i-1][j] + grid[i][j-1]
*/
int col = obstacleGrid.Length;
int row = obstacleGrid[0].Length;
// If the first or the last cell are obstacle, we don't need to run the algorithm.
if (obstacleGrid[col-1][row-1] == 1 || obstacleGrid[0][0] == 1){
return 0;
}
// If the grid only have 1 row or colum and there is no obstacle, return 1 else 0.
if (col == 1 || row == 1){
// Check the column
if (obstacleGrid[0].Contains(1)){
return 0;
}
// Check the row
for (int i = 0; i < row; i++){
if (obstacleGrid[0][i] == 1) {
return 0;
}
}
return 1;
}
/*
Fill up the first row and column with 1s unless there is an obstacle which we will
flip the value of that to 0 as we encouter them.
Any cell on the right of an obstacle in the first row will have a value of 0.
Any cell below an obstacle in the first column will have a value of 0.
*/
for (int i = 0; i < col; i++){
// There is alredy a blocker to the left of the current cell
if (i > 0 && obstacleGrid[i-1][0] == 0){
obstacleGrid[i][0] = 0;
}
// This cell is a blocker
else if (obstacleGrid[i][0] != 1) {
obstacleGrid[i][0] = 1;
}
else {
obstacleGrid[i][0] = 0;
}
}
for (int i = 1; i < row; i++){
// There is alredy a blocker above the current cell
if (i > 0 && obstacleGrid[0][i-1] == 0){
obstacleGrid[0][i] = 0;
}
// This cell is a blocker
else if (obstacleGrid[0][i] != 1) {
obstacleGrid[0][i] = 1;
}
else {
obstacleGrid[0][i] = 0;
}
}
// Looping through every cell in the grid excluding the first row and column.
for (int i = 1; i < col; i++){
for (int j = 1; j < row; j++){
if (obstacleGrid[i][j] == 1){
obstacleGrid[i][j] = 0;
}
else {
obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
}
}
}
return obstacleGrid[col-1][row-1];
}
}Also read,